This problem involves solving the heat conduction equation:

$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$

Subject to the following conditions:

1. Boundary Conditions: $u(x,t) = 0 \quad \text{at} \quad x = 0 \quad \text{and} \quad x = 1 \quad \text{for} \quad t > 0$ This implies that the temperature at the boundaries is fixed at 0.

2. Initial Condition: $u(x,0) = \sin(\pi x) \quad \text{for} \quad 0 \leq x \leq 1$ This specifies the temperature distribution along the rod at time $t = 0$ as $\sin(\pi x)$.

---

Solution Outline:

The problem is solved using the method of separation of variables. Here’s the detailed explanation for part (ii):

---

1. Assume a Separable Solution:

We assume a solution of the form: $u(x, t) = X(x)T(t)$ where $X(x)$ is a function of $x$ only, and $T(t)$ is a function of $t$ only.

---

2. Substitute into the Heat Equation:

Substituting $u(x,t) = X(x)T(t)$ into $\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}$: $X(x) \frac{dT(t)}{dt} = T(t) \frac{d^2X(x)}{dx^2}$

Divide through by $X(x)T(t)$ (assuming $X(x), T(t) \neq 0$): $\frac{1}{T(t)} \frac{dT(t)}{dt} = \frac{1}{X(x)} \frac{d^2X(x)}{dx^2} = -\lambda$

Here, $-\lambda$ is the separation constant.

---

3. Solve for $X(x)$:

The spatial equation becomes: $\frac{d^2X(x)}{dx^2} + \lambda X(x) = 0$

The boundary conditions are $X(0) = 0$ and $X(1) = 0$.

The general solution is: $X(x) = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} x)$

Applying the boundary conditions:

• At $x = 0$, $X(0) = B = 0$.
• At $x = 1$, $X(1) = A \sin(\sqrt{\lambda}) = 0$.

For $A \neq 0$, $\sin(\sqrt{\lambda}) = 0$, which implies: $\sqrt{\lambda} = n\pi, \quad n = 1, 2, 3, \dots$

Thus, $\lambda = n^2 \pi^2$, and the solutions for $X(x)$ are: $X_n(x) = \sin(n\pi x)$

---

4. Solve for $T(t)$:

The temporal equation becomes: $\frac{dT(t)}{dt} + \lambda T(t) = 0$

Substituting $\lambda = n^2 \pi^2$: $\frac{dT(t)}{dt} + n^2 \pi^2 T(t) = 0$

This is a first-order linear differential equation with the solution: $T_n(t) = C_n e^{-n^2 \pi^2 t}$

---

5. Combine Solutions:

The general solution is: $u(x, t) = \sum_{n=1}^\infty C_n \sin(n\pi x) e^{-n^2 \pi^2 t}$

---

6. Apply Initial Condition:

At $t = 0$: $u(x, 0) = \sum_{n=1}^\infty C_n \sin(n\pi x) = \sin(\pi x)$

Using Fourier series, the coefficients $C_n$ can be determined. Comparing terms: $C_1 = 1, \quad C_n = 0 \quad \text{for} \quad n \neq 1$

Thus, the solution simplifies to: $u(x, t) = \sin(\pi x) e^{-\pi^2 t}$

---

Final Answer:

$u(x, t) = \sin(\pi x) e^{-\pi^2 t}$

This describes the temperature distribution over time, starting with the initial distribution $\sin(\pi x)$, and eventually decaying due to the heat equation.

---

Let me know if you want further clarification! Here are related questions to expand your understanding:

1. How do boundary conditions influence the solution of the heat equation?
2. Why is the separation constant $\lambda$ negative in this problem?
3. What happens to the solution as $t \to \infty$?
4. Can the same method be applied for a different initial condition, e.g., $u(x, 0) = \sin(2\pi x)$?
5. How does the solution change for a rod of length $L$ instead of $1$?

Tip: Mastering separation of variables involves recognizing the interplay between boundary conditions and eigenfunctions like sine and cosine.